1. Feb 6, 2005

the_d

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A jogger runs in a straight line with an average
velocity of 4:7 m/s for 5.1 min, and then with
an average velocity of 4.1 m/s for 3 min.
What is her total displacement? Answer in
units of meters

2. Feb 6, 2005

christinono

use the formula:
$$d = V_{average}t$$

3. Feb 6, 2005

the_d

and then I subtract the two to find the final displacement

4. Feb 6, 2005

dextercioby

No,you have to add the 2 distances.

Daniel.

5. Feb 6, 2005

christinono

Well, it all depends what the direction is. If he runs in the same direction both times, you have to add the 2 displacements to find the total displacement.

6. Feb 6, 2005

thanks dude

7. Feb 6, 2005

dextercioby

Even if the runs in the opposite direction on the second time interval,the TOTAL distance would still be gotten by adding the 2.

Daniel.

8. Feb 6, 2005

christinono

Aren't the words "distance travelled" and "displacement" different. I though displacement meant distance and direction from the original starting point.

9. Feb 6, 2005

the_d

for the same question, to get the avg. velocity during the time what should i do? i know it is not just to add the 2 velocities given and divide by 2

10. Feb 6, 2005

christinono

Average velocity is total distance divided by total time.

11. Feb 6, 2005

dextercioby

I think that u've given yourself the answer. Besides,since in the problem we're not given the directions (so that we would think it as a vector projection problem),it's natural to assume the simpler version,which is what the teacher would want,right...? :tongue2:

Daniel.

12. Feb 6, 2005

the_d

new question

A 500-kilogram sports car accelerates uni-
formly from rest, reaching a speed of 30 me-
ters per second in 6 seconds.
During the 6 seconds, the car has traveled
a distance of how many meters??

i got 180m is that correct?
i used : distance = avg. velocity * change in time
= (30m.s) * 6
= 180m

Last edited: Feb 6, 2005
13. Feb 6, 2005

christinono

ha ha, very funny :tongue:

14. Feb 6, 2005

dextercioby

Daniel.

15. Feb 6, 2005

christinono

Careful... 30m/s is not the average velocity, but rather the final velocity. Remember, average velocity is total displacement divided by the total time, or in this case (if a is constant),
$$\frac{V_i + V_f}{2} = \frac{0+30m/s}{2}$$

16. Feb 6, 2005

the_d

where did u get the 2 from???

17. Feb 6, 2005

christinono

Isn't the average the sum of the 2 velocities divided by 2? In this case, the displacement would be: $$d = V_{ave}t$$

18. Feb 6, 2005

dextercioby

Kay,guys,let's leave side talks and refer to the initial problem ...

Daniel.

19. Feb 6, 2005

the_d

your saying I take 30m/s and divide it by 2?? that is not right either.

20. Feb 6, 2005

christinono

I's not "side talking". I'm referring to the "new question". Here, average velocity IS involved in finding the distance.

Anyways, gotta go. I hope I didn't irritate you too much Daniel